question_answer

The angle between two diagonals of a cube is

A)  \[\frac{\pi }{3}\]

B)  \[{{\cos }^{-1}}\left( \frac{1}{3} \right)\]

C)  \[{{\cos }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]

D)  None of these

Correct Answer: B

Solution :

 Let side of cube = a unit The diagonals of a cube are AL and OF. \[\therefore \]Direction ratios of AL are\[a,-a,-a\]and of OP are a, a, a. \[\therefore \] \[\cos \theta -\frac{|{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\] \[=\frac{|{{a}^{2}}-{{a}^{2}}-{{a}^{2}}|}{a\sqrt{3}.a\sqrt{3}}=\frac{1}{3}\] \[\Rightarrow \] \[\theta ={{\cos }^{-1}}\left( \frac{1}{3} \right)\]